Thermionic Emission

3: Thermionic Emission Purpose While we think of quantum mechanics being best demonstrated in processes that show discontinuous change, historically quantum mechanics was first revealed in systems where a large number of particles washed out the jumps: blackbody radiation and thermionic emission In this lab you will investigate these two phenomena in addition to classical space-charge limited electron emission: Child’s Law Introduction Metals, as demonstrated by their ability to conduct an electric current, contain mobile electrons (Most electrons in metals, particularly the “core” electrons closest to the nucleus, are tightly bound to individual atoms; it is only the outermost “valence” electrons that are somewhat “free”.) These free electrons are generally confined to the bulk of the metal As you learned in E&M, an electron attempting to leave a conductor experiences a strong force attracting it back towards the conductor due to an image charge: Fx = − e2 4πǫ0 (2x)2 (3.1) where x is the distance the electron is from the interface and e is the absolute value of the charge on an electron Of course, inside the metal the electric field is zero so an electron there experiences zero (average) force You can think of these valence electrons as bouncing around inside a box whose “walls” are provided by the image-charge force (Odd to think: the “walls” are non-material force fields; the “inside” of the box is filled with solid metal.) Since temperature is a measure of random kinetic energy, if we increase the temperature of the metal, the electrons will be moving faster and some will have enough energy to overcome the image-charge force (which after all becomes arbitrarily small at large distances from the interface) and escape This is electron “evaporation” The higher the temperature the larger the current of escaping electrons This temperature induced electron flow is called thermionic emission Starting in 1901, Owen Richardson studied this phenomenon and in 1929 he received the Nobel prize in Physics for his work A hot wire will be surrounded by evaporated electrons An electric force can pull these electrons away from the wire — the larger the electric force, the larger the resulting current of electrons The precise relationship between the voltage and the resulting current flow 41 42 Thermionic Emission electric current density JA V=0 VA cathode vacuum anode accelerating electrons metal x x=0 x=b Figure 3.1: A planar cathode and a planar anode are separated by a distance b A positive potential difference VA attracts electrons from the cathode to the anode, so the speed of the electrons v(x) increases as they approach the anode The moving electrons constitute an electric current from anode to cathode The resulting steady current density is called JA is called Child’s law1 (or the Child-Langmuir law, including Langmuir who independently discovered it while working at G.E.) In this experiment you will measure both Child’s Law and the Richardson Effect Child’s Law Consider a planar interface between a metal (x < 0) and “vacuum” (x > 0) Vacuum is in quotes because this region will contain escaped electrons—a ‘space charge’—rather than being totally empty2 The number of electrons per volume (i.e., the number density) is denoted by n In this experiment, the metal will be heated (i.e., its a ‘hot cathode’ or filament) which will result in a supply of electrons ‘evaporated’ from the metal into the vacuum An additional conducting sheet (the anode) is located at x = b A positive potential difference, VA , between the cathode and the anode plane provides a force pulling these electrons from the vicinity of the cathode towards the anode The result is a stream of moving electrons (a current); the number density n(x) and speed v(x) of these electrons will depend on location, x, between the plates The negatively charged electrons moving to the right constitute a steady electric current density to the left, i.e., a steady conventional electric current from the anode to the cathode: J = −en(x)v(x) = −JA (3.2) Since the electrons leave the metal with (nearly) zero speed at zero potential, we can calculate their speed along the path to the anode using conservation of energy: mv − eV (x) = (3.3) v= 2e V (x) m (3.4) Clement Dexter Child (1868–1933) Born: Madison, Ohio, A.B Rochester, Ph.D Cornell In fact a perfect vacuum is not possible, so the word “vacuum” actually refers simply to a region with relatively few particles per volume 43 Thermionic Emission where V (x) is the potential difference (“voltage”) at x and m is the mass of an electron Because the accelerating electrons constitute a steady current (i.e., JA doesn’t depend on position), n(x) must decrease as the electrons speed toward the anode The varying space charge density affects the electric potential in the “vacuum” according to Poisson’s equation3 : ∂2V ρ(x) en(x) =− = (3.5) ∂x ǫ0 ǫ0 Putting these pieces together with have the differential equation: d2 V JA = = dx ǫ0 v(x) ǫ0 JA 2e m (3.6) V (x) Since the electric field will be zero at the interface, we have a pair of initial conditions: ∂V = ∂x x=0 V |x=0 = (3.7) (3.8) This differential equation looks a bit like Newton’s second law: d2 x F (x(t)) = dt m (3.9) as you can see if in Newton’s second law you substitute: t −→ x x(t) −→ V (x) JA F (x(t)) −→ m ǫ0 2e m V (x) Recall that force problems are often most simply solved using conservation of energy and that conservation of energy was proved using an integrating factor of dx/dt If we try the analogous trick on our voltage problem, we’ll multiply Poisson’s equation by dV /dx: dV d2 V × dx dx2 dV dx JA = 2e m ǫ0 ′ JA = V (3.10) ′ (3.11) + constant 2e m ǫ0 dV dx 2e m ǫ0 V JA = dV dx V −2 × (3.12) The initial conditions require the constant to be zero, so dV dx JA = ǫ0 V 2e m Poisson’s equation is derived in the Appendix to this lab 2 (3.13) 44 Thermionic Emission or 4JA dV = dx V 2e m ǫ0 (3.14) This differential equation is separable: dV V V 4JA = 2e m ǫ0 4JA = 2e m ǫ0 dx (3.15) x (3.16) where again the initial conditions require the constant of integration to be zero Finally: 2  9JA   x3 V (x) = (3.17) 2e 4ǫ0 m Of course, V (b) is the anode voltage VA , so we can rearrange this equation to show Child’s law: 2e 4ǫ0 (3.18) V JA = A 9b2 m Much of Child’s law is just the result of dimensional analysis, i.e., seeking any possible dimensionally correct formula for JA Our differential equation just involves the following constants with dimensions (units) as shown: b : L E M L2 /T VA : = Q Q ǫ0 (3.19) (3.20) Q2 Q2 Q = EL M M L3 /T Q/T L2 2e ≡k : m JA : (3.21) (3.22) where the dimensions are: L=length, T =time, M =mass, E=energy, and Q=charge To make a dimensionally correct formula for JA , we just eliminate the M dimension which we can only with the combination: VA k : Q3 (3.23) T3 We can then get the right units for JA with: VA k b2 = k 32 V b2 A : Q/T L2 (3.24) Thus the only possible dimensionally correct formula is JA ∝ k 32 V b2 A (3.25) 45 Thermionic Emission cathode: hot filament, radius a anode anode l cathode b Figure 3.2: Coaxial cylinders: an inner wire (radius a) and outer cylindrical anode (radius b), form a vacuum tube diode The cathode is heated so electron evaporation is possible, and a potential difference VA attracts electrons from the cathode to the anode The speed of the electrons v(r) increases as they approach the anode The moving electrons constitute a steady electric current from anode to cathode Since the same current is spread out over larger areas, the current density, J, between the cylinders must be proportional to 1/r The exact proportionality constant, found from the differential equation, is (as usual) is not hugely different from We have derived Child’s law for the case of infinite parallel plates, but you will be testing it in (finite length) coaxial cylinders The inner wire (radius a) is the cathode; the outer cylinder (radius b) is the anode Your cylinder with have some length ℓ, but we will below consider infinite length coaxial cylinders Note that dimensional considerations require that the anode current per length should be given by a formula like: I/ℓ ≡ j ∝ k 23 V b A (3.26) although we could have an arbitrary function of the radius ratio: b/a on the right-hand-side From Poisson’s equation4 we have: ∇2 V = J I j = = ǫ0 v(r) 2πrℓǫ0 v(r) 2πrǫ0 2e m V −2 (3.27) Using the Laplacian in cylindrical coordinates we find: ∂2V j ∂V = + ∂r r ∂r 2πrǫ0 2e m V −2 (3.28) There is no known formula for the solution to this differential equation, but we can make considerable progress by writing the differential equation in terms of dimensionless quanti4 Poisson’s equation is derived in the Appendix to this lab 46 Thermionic Emission ties: r/a = ρ  V =  (3.29) 2 ja 2e m 2πǫ0  f (ρ) (3.30) yielding: 1 ∂f 1 ∂2f + = f ′′ (ρ) + f ′ (ρ) = f − 2 ∂ρ ρ ∂ρ ρ ρ (3.31) f (1) = (3.32) with initial conditions: ′ f (1) = (3.33) We can numerically solve this differential equation using Mathematica: NDSolve[{f’’[p]+f’[p]/p==1/(p Sqrt[f[p]])}, f[1]==0, f’[1]==0, {f},{p,1,200}] It’s actually not quite that simple The cathode, at ρ = 1, is actually a singular point of the differential equation (i.e., f ′′ (1) = ∞) However the situation very near the cathode is well approximated by the planar case, where we’ve shown:  2  3 9I 9JA   V (x) =  x3 = 4ǫ0 2e 2πaℓ4ǫ0 m 2   r−a ja  = a 2e 2πǫ 2 2e m 4   (r − a) =  2 9ja 2π4ǫ0 2e m  r−a a (3.34) m So, near the cathode (i.e., ρ slightly larger than 1): f (ρ) ≈ 4 (ρ − 1) (3.35) We can use this approximation to start our numerical differential equation solution at a non-singular point (like ρ = 1.00001) Real devices are designed with b/a ≫ The behavior of f for large ρ can be determined by finding A and α for which f = Aρα is a solution to the differential equation One finds: f= ρ (3.36) A useful approximation for the range: 100 < b/a < 1000 is: f= ρ +2 (3.37) 47 Thermionic Emission f(ρ) for ρ→1 10 f(ρ) for ρ→∞ 60 f 50 40 30 f(ρ) exact 20 10 50 100 ρ 150 200 10 12 ρ 14 Figure 3.3: The plot on the left displays the dimensionless voltage f obtained by numerical solution to the differential equation The plot on the right compares various approximations for f to this numerical solution (For example, the device used in lab has b/a = 121.5 For this value, the differential equation gives f = 44.136; the above approximation gives: f = 44.130.) We recover Child’s law by rearranging (3.30): 2πǫ0 2e m VA f (b/a) a = j = I/ℓ (3.38) Note: Langmuir’s original work (Phys Rev 22, 347 (1923)) on this subject is expressed in terms of β where:  −→  ρ→1 (ρ − 1)2 f β (ρ) ≡ = (3.39)  −→ ρ ρ→∞ So: 8πǫ0 ℓ 9bβ 2e m VA2 = I (3.40) β = 1.072 for the device used in lab Richardson’s Law Most any thermal process is governed by the Boltzmann factor: exp − ∆E kT = e−∆E/kT (3.41) where k is the Boltzmann constant Approximately speaking the Boltzmann factor expresses the relative probability for an event requiring energy ∆E in a system at (absolute) temperature T Clearly if ∆E ≫ kT , the probability of the event happening is low If an electron requires an energy W (called the work function) to escape from the metal, The 48 Thermionic Emission Boltzmann factor suggests that this would happen with relative probability e−W/kT Thus you should expect that the current emitted by a heated metal would follow: I ∼ e−W/kT (3.42) Clearly you should expect different elements to have different work functions, just as different atoms have different ionization potentials What is surprising is that the proportionality factor in the above equation includes a universal constant — that is, a constant that just depends on the properties of electrons (and, most importantly, Planck’s constant, h) and does not depend on the type of material (This situation is similar to that of blackbody radiation, in which photons rather than electrons are leaving a heated body, which was Planck’s topic in discovering his constant We will take up this topic on page 53.) Thermionic emission probes the quantum state of the electrons statistically, whereas the photoelectric effect probes much the same physics electron by electron (The next level problem is to explain why this universal constant (the Richardson constant, A) in fact does depend a bit on the material.) To show: J = AT e−W/kT (3.43) where A= 4πemk2 = 1.2 × 106 A/m2 K2 h3 (3.44) Quantum Theory: Free Electron Gas Instead of thinking about electron particles bouncing around inside a box, de Broglie invites us to consider standing waves of electron probability amplitude: ψ = N exp(ikx x) exp(iky y) exp(ikz z) = N eik·r (3.45) Recall that vector k is the momentum, p = mv, of the electron and = h/2π Periodic boundary conditions on the box (which we take to be a cube with one corner at the origin and the diagonally opposite corner at the point r = (L, L, L)) require each component ki to satisfy: 2π ni ki = (3.46) L where each ni is an integer Thus each wave function is specified by a triplet of integers: n = (nx , ny , nz ), the n-vector Applying Schr¨ odinger’s equation, we find that this wavefunction has energy: k2 (2π )2 (n2x + n2y + n2z ) (2π )2 n2 = = (3.47) E(n) = 2m 2mL2 2mL2 Notice that there is a definite relationship between the velocity vector of the electron and the n-vector 2π n (3.48) v= mL Another way of saying the same thing is that allowed quantum-state velocities form a cubic lattice with cube-side 2π /mL The number of states with electron velocities in some specified region (for example a velocity-space parallelepiped with sides: ∆vx ∆vy ∆vz ) can For a review see: http://britneyspears.ac/physics/dos/dos.htm 49 Thermionic Emission vacuum metal vacuum image−charge potential  unoccupied   levels U W   occupied  levels EF zero−force (flat) potential inside metal Figure 3.4: Electrons in the metal experience a constant confining potential of depth U Possible quantum mechanical states for these electrons are displayed as horizontal lines Electrons fill all the available states up to the Fermi energy, EF The work function, W , is defined at the minimum energy needed to remove an electron from the metal As shown above: W = U − EF be found from the number of 2π /mL sided cubes that fit into the region, which is the volume of that velocity-space region divided by (2π /mL)3 Hence: ∆vx ∆vy ∆vz (3.49) (2π /mL)3 ∆vx ∆vy ∆vz number of states per volume with velocity between v and v + ∆v = (2π /m)3 m ∆vx ∆vy ∆vz = N ∆vx ∆vy ∆vz (3.50) = 2π number of states with velocity between v and v + ∆v = where N is the (constant) density of states in velocity space Quantum Theory: Fermi Energy Fermions (half-integer spin particles), in contrast to bosons (integer spin particles), cannot group together Since the electron is “spin 21 ”, each of the above states can hold at most electrons: one spin up and one spin down The probability that a particular fermion state with energy E will be occupied is given by a generalization of the Boltzmann factor called Fermi-Dirac statistics: (3.51) f (E) = F + exp E−E kT where EF is called the Fermi energy The Fermi energy is basically a disguise for the number of electrons, as, approximately speaking, it is the dividing line between occupied states and unoccupied states (If the Fermi energy is high, there must be lots of occupied states and hence lots of electrons.) Note that if E ≫ EF , the exponential factor is huge and we can 50 Thermionic Emission v x ∆t A Figure 3.5: Consider just those electrons with some particular x-velocity, vx In order to hit the wall during the coming interval ∆t, an electron must be sufficiently close to the wall: within vx ∆t The number of such electrons that will hit an area A will be equal to the number of such electrons in the shaded volume (which is the perpendicular extension of A a distance vx ∆t into the volume) Note that many electrons in that volume will not hit A because of large perpendicular velocities, but there will be matching electrons in neighboring volumes which will hit A To find the total number of hits, integrate over all possible vx neglect the “+1” in the denominator so f (E) ≈ exp − E − EF kT (3.52) that is, if E ≫ EF Fermi-Dirac statistics approximate the Boltzmann factor Classical Theory: Electron Escape The density of states combined with the Boltzmann factor gives us the number of free electrons per unit volume with a particular velocity In order for an electron to escape during some time ∆t, it must have vx sufficient to overcome the image-charge barrier and it must be sufficiently close to the wall All the electrons with vx > 2U/m within a distance vx ∆t, will escape, where U is the depth of the potential well for the electrons Thus the number of electrons escaping through area A during ∆t is: ∞ ∞ √ dvx 2U/m ∞ dvy −∞ −∞ ∞ = 2N eEF /kT A∆t √ = 4πm(kT )2 (2π )3 dvz 2N f (E) Avx ∆t ∞ ∞ 2U/m A∆t exp − 2 e−mvz /2kT dvz e−mvy /2kT dvy e−mvx /2kT vx dvx −∞ (U − EF ) kT −∞ (3.53) where we have used the Gaussian integral: ∞ e−αz dz = −∞ π α (3.54) Thermionic Emission 55 be determined At room temperature, the filament resistance, RW , can be calculated from the filament geometry (see Table 3.1) and the resistivity of W-filament material at room temperature: ρ293 = 5.49 µΩ · cm (You should calculate: RW ∼ Ω.) Then: Rsupport = Rmeasured − RW (3.63) Because room temperature Rmeasured is ‘small’ (and hence error prone), you will make three distinct measurements of it Begin by sourcing mA and then 10 mA into the room temperature filament (using the 2420), reading the resulting voltages on the 192, and calculating the corresponding Rmeasured Follow up those two measurements with a fourterminal resistance measurement just using the 192 (If Rmeasured is substantially larger than ∼ Ω, confirm that you have firm, low-resistance contacts between the socket and the tube Working the socket-tube contact may be required to find the lowest resistance spot.) Maximum Filament Current, 2420 Voltage Compliance Limit Following your determination of Rsupport in a room temperature tube, check out the conditions required to stay just below the tube’s maximum ratings (4.75 V, 2.5 A) Using the 2420, successively source filament currents of 2.0 A, 2.1 A, 2.2 A, to directly determine the maximum current you can use without exceeding the 4.75 V limit across the tube’s filament Note that at just below maximum conditions, the 2420 will probably need to produce voltages above 4.75 V (because of the resistance of the external wires: the voltage drop across the connecting wires is not zero) Record the maximum 2420 voltage and tube current allowed by the tube’s ratings; you will need these numbers in step #2 of your computer program Data Collection Plan You will be collecting two types of data at the same time: thermal characteristics of the filament and the thermionic properties of the tube (anode current/voltage relationship) Starting at a filament current of 0.9 A, increase the current flowing through the filament in steps of 0.1 A to a maximum current (found as described above, about 2.4 A) and then reverse those steps down to a filament current 1.0 A The up-sweep in filament current followed by the down-sweep will allow you to test for hysteresis At each step in current, allow the filament to approach thermal equilibrium (wait, say, 15 seconds) and then measure the voltage across and current through the cathode/anode Calculate filament temperature two ways (Equations (3.56) and (3.62)) Average the two to estimate the temperature, and use half the absolute value of the difference to estimate the uncertainty You see above a classic example of systematic error The temperature is measured two different ways Direct application of error propagation formulas to these temperatures calculated from 6-digit meter values would suggest small uncertainties However the two temperatures in fact disagree If only one method had been used to measure temperature, we would have badly underestimated the error 56 Thermionic Emission T vs Power: Testing Stefan-Boltzmann By conservation of energy we know that the power dumped into the filament (mostly from electrical heating, but also from other sources like radiation from the room temperature environment to the filament) should equal the power out of the filament (from black-body radiation and other factors like conduction down the supports) Thus: ǫT σAT = I RW + constant I RW + constant T4 = ǫT σA y = bx + a (3.64) (3.65) (3.66) A graph of T vs power should be a straight line from which you will determine ǫT (Note that the error in power is quite small, so we have properly used it as an x variable.) In order to test this relationship you will want to make a file containing the filament power, T (use the average of the two temperatures: (Ti4 + Tr4 )/2), and the error in T (use half the difference from the two temperatures: |Ti4 − Tr4 |/2) IA vs VA : Testing Child and Richardson You will collect anode current vs voltage curves for each different filament (cathode) temperature Use the Keithley 2400 to sweep the anode voltage logarithmically from V to 120 V (Note the maximum limits for the anode: 0.055 A or 125 V Do not exceed either!) According to Child’s law, the anode current, IA , should increase in proportion to VA2 Of course, at sufficiently high voltage the current will be limited by the maximum electron evaporation rate, and a current plateau forms at a level given by Richardson’s law At the maximum filament current (corresponding to the maximum filament temperature and evaporation rate), plateau formation occurs at very high voltage and you have the longest run of data following Child’s law Make a file containing VA , IA , and δIA which you can use to fit to the Child’s law functional form: IA = k1 (VA − k2 ) (3.67) In addition, you will want to make a big continuous file containing: VA , IA at every temperature tested The current plateaus can be extracted from this file and fit to the Richardson relationship: IA = k1 AT e−k2 /T (3.68) Computer Data Collection As part of this experiment you will write a computer program to control the experiment Plagiarism Warning: like all lab work, this program is to be your own work! Programs strikingly similar to previous programs will alarm the grader I understand that this will often be a difficult and new experience Please consult with me as you write the program, and test the program (with tube disconnected!) before attempting a final data-collecting run Your program will control all aspects of data collection In particular it will: Thermionic Emission 57 Declare and define all variables Open (i.e., create integer nicknames—i.e., iunit—for) the enets gpib0 and gpb1 Initialize meters—requires knowing the gpib primary address—i.e., iadd—of the meter and the iunit it is attached to Get the status of each meter after you have initialized it (a) Each source-meter must be told the maximum voltage and current it must produce during the experiment Initialize the 2400 (anode voltage/current) for the tube maximum ratings (b) Initialize the 2420 (filament voltage/current) for the near tube-maximum ratings found above11 In the following I assume the current maximum is 2.4 A, but it may be different for your tube (c) Initialize the 192 for autorange DC voltage measurements Open the files: (a) filament.dat (intended for: If , Vf , Tr , Ti of filament) (b) stefanB.dat (intended for: power, T , δT of filament) (c) VI.dat (intended for: all VA , IA of anode, with comments (‘!’) for filament If , Tr , Ti ) (d) child.dat (intended for: VA , IA , δIA of anode at maximum filament current) (e) child-.dat (like above but intended for a downsweep of anode voltage) (f) rich.dat (intended for: Ti , Tr , IA — i.e., the estimated temperatures and the corresponding maximum anode current for Richardson’s Law) Tell the 2420 source-meter to source a filament current of 0.9 A Let the system sleep for 60 seconds to approach thermal equilibrium Do a sequence of measurements where the filament temperature is sequentially increased (i.e., a temperature up-sweep) with filament currents ranging from 0.9 A to some maximum (e.g., 2.4 A) current in steps of 0.1 A (a) Tell the 2420 source-meter to source a filament current (If ) (b) Let the system sleep for 15 seconds to approach thermal equilibrium (c) Request a logarithmic sweep of the anode voltage (controlled by the 2400 sourcemeter) from V to 120 V Receive the resulting arrays: VA and IA (d) Turn off the anode voltage (e) Repeat (a) thus receiving an updated version of the filament current (it will be very close to the requested current) (f) Read the 192 to get the filament voltage (Vf ) (g) Using Eqs (3.56) and (3.57), calculate Tr based on the calculated tube resistance Rmeasured , the calculated room temperature filament resistance and Rsupport (h) Calculate Ti from Eq (3.62) 11 See Hands-on Electrical Measurements, p 54 Recall: the 2420 maximum voltage will need to be a bit above 4.75 V If you have not yet completed those measurements, temporarily initialize with 4.75 V 58 Thermionic Emission (i) Write a line to the file filament.dat reporting: If , Vf , Tr , Ti (j) Write a line to the file stefanB.dat reporting: filament power (RW If2 ), T , and δT (see p 56) (k) Write a comment line (i.e., starts with ‘!’) to the file VI.dat reporting filament data: If , Tr , Ti (l) Write the anode sweep data (one data-pair per line) in the file VI.dat (m) Write a line to the file rich.dat reporting Ti , Tr , IA Use IA at VA =120 V as the estimated plateau current (When the experiment is over, you will need estimate δIA based on hysteresis, and may need to delete IA values if, for example, the current did not plateau or if cold emission substantially added to the plateau current.) (n) Increment the filament current by 0.1 A and return to (a) Collect data for Child’s Law Begin by repeating a normal anode sweep at the maximum filament current; follow all the steps (a)–(m) outlined in above In addition, write the anode sweep data (VA , IA , δIA ) in the file child.dat (one data-triplet per line) In this case, δIA will be calculated from the manufacturer’s specs: percent+digits; the fortran function eAk2400 can this error calculation automatically Now check for hysteresis by doing a reverse anode sweep: from 120 V down to V Write this reverse anode sweep data (VA , IA , δIA ) in the file child-.dat Do a sequence of measurements where the filament temperature is decreased (i.e., a temperature down-sweep) by sequentially sourcing filament currents from one step down from maximum (e.g., 2.3 A) back to 1.0 A Follow steps (a)–(m) used in the temperature up-sweep (part above) and then: (n) decrement the filament current by 0.1 A and return to (a) Turn off the output of the 2420 10 Close all files Note that the 0.9 A filament current data is just trash collected so the 1.0 A filament current data is taken on a pre-warmed filament Observations ” While the computer collects the data observe the light from the filament (There is a 16 diameter hole in the anode allowing light from the mid-point of the filament to escape.) Initially the filament will not appear to be incandescent (i.e., not a source of light at all: dark) so it may help to turn off the lab lights to better observe the beginning of visible incandescence According to the Stefan-Boltzmann law the light energy radiated depends on T , so small changes in T produce big changes in light intensity In addition to changes in light intensity, you should look for the more subtle change in light color: from a dull red to a brilliant yellow Record your observations! At what filament temperature did you first see the filament producing light? 59 Thermionic Emission FP–400 Vacuum Tube 01 Ia (A) 001 1.E–04 1.E–05 1.E–06 1.E–07 10 Va (V) 100 Figure 3.9: The temperature dependence of thermionic emission in a FP-400 vacuum tube Each curve shows the anode current-voltage relationship at a particular filament temperature At high filament temperatures and low anode voltages the anode current follows Child’s law: an upward sloping straight line on this log-log plot At sufficiently high anode voltage, the filament current plateaus at a value given by Richardson’s law At low filament temperatures and high anode voltages, “cold emission” may augment (distort) the plateau Data Analysis Beginnings The main result of this experiment is a plot similar to Figure 3.9 showing the anode currentvoltage relationship at various filament temperatures Production of such a multi-plot is a bit complex, and you will almost certainly need further detailed instructions from your instructor on using the program Nplot Note that since this is a log-log plot, negative anode currents must also be edited out of VI.dat to make this plot It may be difficult to determine the plateau level for the lowest and highest filament temperatures, in which case you must edit out those data points in rich.dat Note: If the high-temperature V I sweep reaches a plateau, then Child’s Law will not apply at high VA so child.dat will require editing If it does not reach a plateau, then Richardson’s Law does not apply so rich.dat will require editing In Figure 3.9, I see no sign of a plateau at the maximum filament current, but in the range If = 1.1 to 2.3 A, plateau currents can be determined (I.e., I removed from rich.dat the non-plateau data point for If = 2.4 A, and also the aberrant initial data for If = 1.0 A.) Additionally a glance at the file rich.dat should demonstrate systematic temperature error: Ti = Tr Clearly fits assuming T = Tr will produce different results from fits assuming T = Ti Absent further information, our best estimate for T must be something like the 60 Thermionic Emission Richardson’s Law Anode Current (A) 01 001 1.E–04 1.E–05 2600 2400 2200 2000 1800 Temperature (K) 1600 Figure 3.10: A Richardson Plot of thermionic emission in a FP-400 vacuum tube Each data point displays the plateau anode current at a particular filament temperature The curve through the data fits for the work function; the slightly steeper curve uses the book value for the work function Child’s Law Ia (A) 01 001 0001 10 Va (V) 100 Figure 3.11: A plot of the space-charge limited thermionic emission in a FP-400 vacuum tube (Child’s law) The data was taken at a filament current of 2.4 A Every other data point has been eliminated so the fit line is not obscured Note that the fit line systematically misses the data, sometimes a bit high others a bit low The measurement errors are tiny, so these small misses result in a “too-large” χ2 Nevertheless, the law provides an excellent summary of the data over a huge range of variation 61 Thermionic Emission Stefan–Boltzmann Law 4.E+13 T^4 (K^4) 3.E+13 2.E+13 1.E+13 Power (W) 10 Figure 3.12: A test of the Stefan-Boltzmann law: that power radiated is proportional to T Note that the fit line hits well within each error bar The χ2 for this fit will be “small” Evidently the average temperature is a better measure of temperature than you might expect from the deviations between Ti and Tr average of Ti and Tr The difference between fit parameters produced assuming T = Tr and those produced assuming T = Ti will allow use to estimate the systematic error in those parameters Now according to Richardson’s law, these plateau currents should satisfy: I = AAT e−W/kT = k1 AT e−k2 /T (3.69) where A is the tungsten filament surface area However, since we are currently only seeking error estimates, simplified analysis is justified Clearly the largest error is in T , so the standard approach would be to put T on the y-axis and I on the x-axis: the opposite of what is implied by the above equation However, we can’t simply solve the above equation for T without some seemingly huge approximations It turns out that e−k2 /T is the significant factor in the above equation, so we start by ignoring the T and assume: I = a e−b/T (3.70) Now if we take ln of both sides: ln(I) = ln(a) − k2 or T 1 = ln(a)/k2 − ln(I) T k2 (3.71) (3.72) This equation is now in a form12 known to WAPP+ Thus you can quickly WAPP+ (I, Ti ) and (I, Tr ) data from rich.dat (for simplicity assume no x-error and no y-error), and 12 Inverse-Natural Log: 1/y = A + B log(x) 62 Thermionic Emission Work Function Systematic Error Temperature (K) 1000 1500 2000 2500 1.E–08 1.E–06 1.E–04 Anode Current (A) 01 Figure 3.13: Simplified analysis suggests Richardson’s Law data can be fit to the InverseNatural Log relationship of Eq 3.72 The filled-square points use y = Ti , the unfilled-square points use y = Tr Systematically different temperature measurements yield systematically different B (slopes in the above plot) and hence systematically different work functions W = −k/B Note the inexact pairing of the data due to temperature hysteresis determine the range of k2 = −1/B consistent with the differing temperature data Request WAPP+ make a linearized plot of your data with x-scale: log and y-scale: inverse as in Fig 3.13 This should allow you to check for aberrant data points (Usually the high temperature curve has not plateaued, and so the high temperature data point must be removed Occasionally the low temperature curve is also aberrant.) The range for W = k2 k has been found by simplified fits; the best value for W will be found (see following) with a fit to the proper functional form with due account for the uncertainty in current caused by hysteresis and the best estimate for the actual temperature The Stefan-Boltzmann Law is the easiest law to check: you can quickly WAPP+ the data in the file stefanB.dat to produce a plot similar to Fig 3.12 I expect you’ll find a small reduced χ2 due to the large systematic error in temperature, so a bit of additional work will be required to estimate δǫT Child’s Law At sufficiently high filament temperature (and low anode voltage), Child’s law governs the anode current-voltage relationship: IA = 8πǫ0 ℓ 9bβ 2e m VA2 (3.73) (see vicinity of Eq 3.40, page 47, for a definition of symbols) At the highest filament temperature (e.g., filament current of 2.4 A) you have saved the (VA , IA , δIA ) data in the Thermionic Emission 63 file child.dat Now fit and plot this data to the functional form: IA = k1 (VA − k2 ) (3.74) (where k2 represents an offset between ground and the actual average voltage of the filament), producing a result similar to Fig 3.11 Do not be surprised if you get a huge reduced χ2 Find an estimate for k1 error either by a ‘fudged fit’ or a bootstrap Follow exactly the same process for the downsweep data in the file child-.dat Generally you will find that the k1 values for the two sweeps differ by more than computer-based value of δk1 Systematic error (here a type of hysteresis) is responsible Note that the usual reduced χ2 alerts us to a problem, but measuring twice (in different ways) provides an estimate (perhaps still an under-estimate) for δk1 : half the difference between the two values of k1 We expect that k1 = 8πǫ0 ℓ 2e m 9bβ (3.75) so with known geometry, the electron charge-mass ratio e/m can be calculated Since the FP-400 is a finite length cylinder (rather than the infinite cylinder discussed in our theory) use the effective length13 = 0.7 × ℓ as the length of the cylinder But what can we use as errors for the ‘book’ values for b and ℓ? Thermal expansion of the filament should, by itself, change ℓ by about 1% (notice the spring-tensioned support for the filament in the FP-400) Based on the sigfigs in the reported values, I estimate: δ(b/ℓ) ≈ 3% (b/ℓ) (3.76) Calculate e/m and its error Compare to the ‘known’ value (citation required!) Stefan-Boltzmann Law You should have already checked for an approximately linear relationship between electrical power in and T : Power = ǫT σAT (3.77) and found a reduced χ2 indicative of large systematic uncertainty in temperature We now seek an estimate for ǫT (with error) at the highest temperature The best value for ǫT can be found by plugging in our best estimates for T and the measured power (found in the file stefanB.dat), and A (calculated based on the dimensions recorded in Table 3.1) Alternatively ǫT could be calculated based on slope as suggested by Eq 3.66 But how should we incorporate the large systematic errors in T and the unknown systematic error in A? For the surface area A, all we know is the ‘book’ values for the dimensions of the filament Based on the sigfigs in the reported values, I estimate: 13 δA ≈ 10% A (3.78) This effective length corrects for current loss through the ends of the cylinder under space-charge situations A smaller correction should be applied when large anode voltages are designed to sweep up all evaporated electrons, i.e., for Richardson’s Law, where 90% electron collection seems appropriate The details of handling such corrections to theory may be found in reference 64 Thermionic Emission (mostly due to the filament diameter, where a ‘small’ uncertainty leads to a large fractional uncertainty) We can then use the ‘high-low’ method of 191 to estimate the range of possible values for ǫT , given the range of possible values for T and A (assume the error in power is negligible) Richardson-Dushman Law You should have already checked for an approximately exponential-inverse relationship between T and IA , edited out aberrant data (e.g., not yet plateaued), and have a range of possible values (due to systematic error in temperature) for k2 in the expected relationship: IA = k1 AT e−k2 /T (3.79) We now seek a treatment incorporating the hysteresis error in IA and the proper functional form, to obtain the best possible value for k2 We will need to manipulate14 the data in rich.dat to bring together equivalent data from the temperature upsweep and downsweep An easy way to get the data into the gnumeric spreadsheet, is to type it to the screen using the UNIX command15 cat: cat rich.dat You can then copy and paste16 this data into gnumeric Our aim to put together data with the same filament current, i.e., to combine the first line of data with the last; to combine the second line of data with the next-to-last; etc This is easily accomplished by typing the data bottom-to-top to the screen using the UNIX command17 tac: tac rich.dat The results can be copy and pasted next to the previous data so that the data we aim to combine is on the same line The best estimate for T is the average of the four T s (the two Ti should be nearly identical); the best estimate for IA is the average of the two IA s; for δIA use half the difference between the two values of IA (|IA1 − IA2 |/2); for δT use half the difference between Tr and Ti (either pair or average the two differences) The result of this ‘data reduction’ is half as many data points, but with a value for δIA based on hysteresis (The alternative for δIA would be the meter manufacturer’s specifications, and we know they are way to small from analysis of Child’s Law.) Report the relative importance of hysteresis and calibration in temperature uncertainty determination For If = 1.2 A record the difference in Tr due to hysteresis and the difference between Tr and Ti which is a temperature calibration uncertainty Copy and paste this reduced data into a new file, and fit it to Eq 3.79 Produce a plot similar to Fig 3.10; include lines both for your best fit and the ‘book’ values of k1 and k2 A glace at Eq 3.69 on page 56 shows that k1 = A (Richardson’s Constant) and kk2 = W (Work Function) 14 This processing could very easily have been done within the program itself I’ve instead opted to make the program as simple as possible at a cost of additional ‘by-hand’ processing in a spreadsheet 15 from concatenate—commonly this command is used to combine several files 16 Note use of “See two two separators as one”: Alt-e 17 clever or what? Thermionic Emission 65 The work function is an atomic quantity, and it is usually expressed in the atomic scale unit eV18 Calculate the work function from your value of k2 in joules and eV and compare to the book19 value of 4.5 eV The book value for A is 0.72 × 106 A/m2 K2 ; This will serve as a fair initial guess for k1 (required for fitting), but we need a better way to calculate A (and particularly the error in A) based on your data20 Since our temperatures are so uncertain, particularly at the low end, the best estimate for the Richardson constant A comes from substituting the book value of the work function and the plateau (T, I) measurement for the highest valid filament temperature into Eq 3.69 We can then estimate the systematic uncertainty in A by using the ‘high-low’ method of 191 (A+ → I + , T − , A− ) Report Checklist Write an introductory paragraph describing the basic physics behind this experiment For example, why did higher cathode currents produce ever higher plateaus of anode current? (This manual has many pages on these topics; your job is condense this into a few sentences and no equations.) Calculations (no errors) of room temperature RW Measurements (4-wire ohmmeter and direct voltage/current) of Rmeasured at room temperature Calculation of Rsupport Observations of the light intensity and color as a function of filament temperature Data files and computer program: Leave them in your UNIX account Print out a copy of your program, the file filament.dat, and the data you used to fit Richardson’s Law; Tape them into your lab notebook Plots similar to Figures 3.9, 3.10 (with fit curve Eq 3.68 and also the Richardson function using “book” values for k1 and k2 ), 3.11 (with fit curve Eq 3.67), 3.12 (with line Eq 3.66) and 3.13 (two separate WAPP+ plots one with Tr data and the other with Ti data) Note that Figure 3.9 is complex to produce Use a file of Nplot commands and feel free to talk to me about how to go about doing this Carefully note the use of log and inverse scales (which requires positive data—edit to achieve this!) Also include a fit report for each fit curve Experimental values (with error range) for: W (in eV), A, e/m and ǫT Show the missing steps leading to Equations (3.31) and (3.36) Substitute the ρ → approximation (Eq 3.35) into the differential equation Equation (3.31) Show that while we not have an exact solution to the differential equation, the singular parts (i.e., those that approach infinity as ρ → 1) cancel Make a final results table, recording your final results (with proper units and sigfigs) adjacent to the corresponding ‘book’ values Don’t forget to record your tube’s identifying letter! 18 Recall: eV = 1.6022 × 10−19 J is the energy an electron gains in going through a potential difference of V 19 Blakemore, Solid State Physics 20 Recall that when systematic errors dominate computer-based errors are usually some sort of non-sense 66 Thermionic Emission Comment: Classical vs Quantum I said above that the presence of an in Richardson’s constant shows that the process is governed by quantum mechanics It is not quite so simple The evaporation of water is not fundamentally different from that of electrons, while the former (due to the larger mass of a water molecule) is well-approximated by a classical calculation Classical evaporation can be calculated much as quantum evaporation using the Maxwell-Boltzmann speed distribution and the number density (of water molecules) rather than the disguised version of this: Fermi energy (EF2 ∝ number density) We can write the classical rate in terms of the quantum with no visible: EF × quantum flux (3.80) classical flux = √ π kT The different temperature dependence21 for the classical flux (T e−W/kT vs T e−W/kT ) cannot be detected experimentally: the Boltzmann factor overwhelms all other temperature dependencies On the other hand, since EF ≫ kT , the expected classical rate is much larger than the quantum rate This played a role in the mistaken idea that thermionic emission was due to surface contamination: the experimental rate seemed too small to be thermal evaporation On the other hand a more fruitful interpretation of the “low” rate was that only a fraction (∼ kT /EF ) of the electrons were thermally active This connected with other observations (like the “small” specific heat of the electron gas) and provided a link to the idea of a degenerate Fermi gas Comment: Uncertainty Inspection of Figures 3.9-3.12 shows that something “funny” is going on (I can’t say “abnormal” or “unusual” as it is neither.) Figure 3.12 shows the unmistakable signs of “small reduced χ2 ”: The fitted line goes nearly dead-center through all 30 error bars, never coming even close to an edge For error bars sized at one standard deviation (σ), you should expect total misses of the error bar about 1/3 of the time In addition recall that each data point is really a double: the same filament current was sourced as part of a temperature up-sweep and as part of a temperature down-sweep These repeated measurement should also frequently miss by a standard deviation, but here they are so close that the two points often look like just one The answer to this puzzle is that the error bars are not displaying statistical (‘random’) error Instead the temperature was measured two different ways (Ti and Tr ), and the error bar represented the deviation between these two measurement methods When different methods of measurement produce different answers for the same quantity, we have a textbook example of systematic error (in contrast to statistical error) Notice that if we had used the statistical deviation of just one measure of temperature, we would seriously underestimated the error Furthermore since quite accurately measured electrical quantities were used to calculate the temperature (via Equation 3.62 or Equation 3.56), application of the usual error propagation methods would also have produced very small errors in T The full range of our uncertainty in temperature is made evident only by measuring T two different ways (This is typically how systematic errors are detected.) 21 Saul Dushman ( Phys Rev 21, 623–636 (1923)), while working at G.E., provided a general thermodynamic argument for the T dependence and the universal nature of A The resulting relationship is therefore sometimes known as the Richardson-Dushman relationship Thermionic Emission 67 Having detected systematic errors, we should seek an explanation In addition to the problems previously cited (particularly use of book values for filament dimensions and the problems associated with Rsupport ), nonuniform filament temperature due to filament supports may be the problem Koller (p 89) reports the filament temperature 0.5 cm from a support is reduced by 15% (and of course the effect is greater closer to the support) Thermionic emission is reduced a similar amount 1.3 cm from a support Thus the quantity we are seeking (a filament temperature) does not even exist, so it is hardly surprising that different measurements give different results (It turns out the Tr is something like the average temperature; whereas Ti is something like the central temperature.) These effects have been investigated, and Koller gives the theory to correct such measurements, but such considerations are beyond the scope of this experiment Figure 3.9 shows the opposite problem, “large reduced χ2 ”: The fitted line systematically misses almost every error bar In this case, the miss might be called “small”, but the error bar is smaller still Once again, errors were not calculated statistically (manufacturer’s specifications were used), so “reduced χ2 = 1” should not really be expected In this case, my guess is that the problem is with our simplified theory (we assumed: infinite cylinders, no random component to the electron velocity [zero electron temperature], uniform filament [temperature, voltage, emissivity, ], perfect vacuum, no incipient current plateau) We could of course test these hypotheses by further experiments with different tubes, but such work is beyond the scope of this experiment (Indeed I have detected VA -sweep hysteresis; correction for this dramatically reduces reduced chi2 , but not all the way to ∼ 1.) Summary: Very large or very small reduced χ2 suggests significant non-statistical errors, a very common —perhaps even the usual— situation Computer generated errors are some sort of none sense in this circumstance Presumably your theory and/or measurement methods are less than perfect That is, of course, No Surprise Science next requires you to guess and perhaps further investigate which imperfections are the leading source of the problems, i.e., what changes to the experiment would ameliorate the problem References Jones & Langmuir GE Review, The Characteristics of Tungsten Filaments as Functions of Temperature 30 (1927) Part I pp 310–19, Part II pp 354–61, Part III pp 408–12 Melissinos Experiments in Modern Physics, 1966, pp 65–80 Preston & Dietz The Art of Experimental Physics, 1991, pp 141–47, 152–61 Blakemore Solid State Physics, 1974, pp 188–95 Koller The Physics of Electron Tubes, 1937, Ch I, VI, VIII G.E FP-400 Description and Rating FP400.pdf found at http://www.tubedata.org 68 Thermionic Emission thickness: ∆x E(x)    Area: A ∆x E(x+∆x) x Figure 3.14: Gauss’ Law is used to calculate the charge between two plates of area A separated by a distance ∆x Since (by assumption) the potential just depends on x, the electric field is in the x direction and is given by E = −dV /dx Appendix—Poisson’s Equation Equation 3.5 and Equation 3.28 made reference to “Poisson’s Equation”, which is really a topic for Physics 341, rather than part of Physics 200 In this appendix, Poisson’s Equation is derived starting from two Physics 200 results: Gauss’ Law: that the electric flux leaving a region depends just on the charge enclosed by that region: ˆ dA = Qenclosed /ǫ0 E·n (3.81) and the relationship between electric field and electric potential (voltage): Ex = − dV dx (3.82) Poisson’s Equation is a differential equation equivalent to Gauss’ Law It is usually written in terms of the Laplacian (∇2 ), which in turn can most easily be expressed in terms of second derivatives w.r.t x, y, and z: ∇2 V = ∂2V ∂2V ∂2V + + = −ρ/ǫ0 ∂x2 ∂y ∂z (3.83) where ρ is the electric charge density We need Poisson’s Equation only in cases where the electric potential depends on just one variable (x or cylindrical r), which simplifies the required proofs considerably As shown in Figure 3.14, if V is a function of x alone: V (x), we can find the charge between two plates of area A using Gauss’ Law: Qenclosed = ǫ0 A (E(x + ∆x) − E(x)) ≈ ǫ0 A dE ∆x dx (3.84) Thus the charge density between the plates is given by: ρ= ǫ0 A dE Qenclosed dE d2 V dx ∆x = = ǫ0 = −ǫ0 volume A∆x dx dx2 (3.85) 69 Thermionic Emission thickness: ∆r E(r+∆r) l E(r) r+∆r Figure 3.15: Gauss’ Law is used to calculate the charge between two coaxial cylinders of length l separated by a distance ∆r Since (by assumption) the potential just depends on r, the electric field is in the r direction and is given by E = −dV /dr which provides what is needed for Equation 3.5 As shown in Figure 3.15, if V is a function of r alone: V (r), we can find the charge between two coaxial cylinders using Gauss’ Law: Qenclosed = ǫ0 l {2π(r + ∆r)E(r + ∆r) − 2πrE(r)} = ǫ0 l {2πr(E(r + ∆r) − E(r)) + 2π∆rE(r + ∆r)} dE ≈ ǫ0 l 2πr + 2πE(r) ∆r dr (3.86) Thus the charge density between the cylinders is given by: ǫ0 l {2πr dE/dr + 2πE(r)} ∆r Qenclosed = = ǫ0 volume 2πrl∆r d2 V dV = − ǫ0 + dr r dr ρ = which provides what is needed for Equation 3.28 dE + E dr r (3.87) (3.88)